Felix played a game where he rolled five dice, six times. In each of these six turns, he would select a number (1 to 6) and score the total value of that number on the dice. So a roll of 2,2,6,6,6 would score him 4 in the 2's, 18 in the 6's, and 0 anywhere else, although Felix never scored 0 on any turn. He played this game a total of five times, scoring the same total in each game, but he didn't score the same value across a row of his scoring grid. In more than one of his games, he rolled five of more than one number, but these games were not consecutive. In any game where Felix rolled five of an odd number, he didn't roll an odd number of any other odd number. For every game x, the number of 6's rolled was 6-x and for all but one of these games, he rolled five x's. Complete his scoring grid.
G A M E
| 1 | 2 | 3 | 4 | 5 |
---+---+---+---+---+---+
1's| | | | | |
S ---+---+---+---+---+---+
2's| | | | | |
C ---+---+---+---+---+---+
3's| | | | | |
O ---+---+---+---+---+---+
4's| | | | | |
R ---+---+---+---+---+---+
5's| | | | | |
E ---+---+---+---+---+---+
6's| | | | | |
---+---+---+---+---+---+
Hint: this is purely a logic problem. The text is written to confuse, but the statements are unequivocal. First work out what your possible values are to fill the grid in, and then follow the constraints set by the problem until you reach a solution. You can assume that Felix used normal six-sided dice.
For reference, I've taken the text of the question and spilt it up into numbered statements:
From (1) and (2) we have the following possible values for each cell of the scoring grid...
From (3) we have that he played the game five times, and from (4) we have that the score was different in each game for any particular number. So the possible values given above each must appear once and once only. Furthermore, we score the same in each game from (3), so this must be a total of 63 per game (by summing all the possible values and dividing by five).
We can now use (7) to start filling in the grid.
Note that, from (7), one of {5, 10, 15, 20, 25} above is not in the correct position. We can work out which one of these it is though. Looking at (5), we already have two sets of 5 in game 1. So either the 10, the 15, the 20, or the 25 are in the wrong position. Taking them in reverse order...
25: By (5) the 25 cannot move to game 1 (only one game would have more than one set of 5), or to game 2 (in which case the two games would be consecutive). By (6) the 25 cannot move to game 3, since that would be five 3's with an odd number of 5's. So if the 25 isn't in game 5, it must be in game 4. Using the smallest possible values (note an even number of 1's and 3's since we have five 5's) we would have [2, 2, 6, 20, 25, 12] in game 4, totalling 67, not 63. Therefore the 25 must remain in game 5.
20: By (5) the 20 cannot move to game 1 or game 2. If it moved to game 3 (noting the five 3's would require an even number of 1's and 5's) we would have minimum values of [2, 2, 15, 20, 10, 18] totalling 67. If it moved to game 5 (noting the five 5's would require an even number of 1's and 3's) we would have minimum values of [2, 2, 6, 20, 25, 6] totalling 61, so we could possibly have either of [4, 2, 6, 20, 25, 6] or [2, 4, 6, 20, 25, 6] in game 5.
15: By (5) the 15 cannot move to game 1 or game 2. By (6) it cannot move to game 5. If it moved to game 4, we would have minimum values of [2, 2, 15, 20, 10, 12] totalling 61, so we could have [4, 2, 15, 20, 10, 12] or [2, 4, 15, 20, 10, 12] in game 4.
10: By (5) the 10 cannot move to game 1. Minimum values putting it in game 3 (noting the five 3's) would be [2, 10, 15, 4, 10, 18] totalling 59. Minimum values putting it in game 4 would be [1, 10, 3, 20, 5, 12] totalling 51. Minimum values putting it into game 5 (noting the five 5's) would be [2, 10, 6, 4, 25, 6] totalling 53. There are no obvious eliminations to be made.
Looking at the 5's row in game 1, we can either put 10 or 20 there (since we have five of an odd number already). Using the 20, and putting minimum (even) values throughout the rest of that column would give [5, 2, 6, 4, 20, 30] totalling 67. Therefore the 10 in the 5's row must go under game 1, and so therefore must the 20 go in game 3. The only solutions allowing the 15 in the 3's row to move require putting the 10 of the 5's in game 4, so the 15 of the 3's must remain in game 3. Filling in the rest of game 1, the 3's can either take the value 6 or 12. If we use 12, the only solution possible is [5, 2, 12, 4, 10, 30]. Using 6 we have either [5, 4, 6, 8, 10, 30] or [5, 8, 6, 4, 10, 30].
Let us now assume that the 20 in the 4's row is not in game 4. If this is the case, comparing possible values for games 1 and 5, we would get the following partial solution:
Using the minimum remaining even values, in game 3 we would then have [4, 6, 15, 8, 20, 18], totalling 71. So the 20 in the 4's row must remain in game 4, and so by process of elimination it must be the 10 in the 2's row that is not in game 2. The correct solution so far:
Looking at game 3, using any value in the 4's higher than 4 will give a sum for that game greater than 63. So game 3 has one of two solutions: [2, 4, 15, 4, 20, 18] or [4, 2, 15, 4, 20, 18]. This further limits the possible values game 1 can take to just one: [5, 4, 6, 8, 10, 30]. And this in turn sets game 3 to be [4, 2, 15, 4, 20, 18]. Furthermore, since game 5 contains five 5's, there can only be an even number of 1's and 3's. The remaining values are 2 in 1's and 12 in 3's, giving the only solution for game 5 as being [2, 6, 12, 12, 25, 6]. The solution now looks like this:
The 10 in 2's has to go in game 4, and by elimination game 2 has 8 in 2's and 16 in 4's:
Putting 15 in 5's for game 2 will give a total greater than 63, so in 5's we have 5 in game 2 and 15 in game 4. This only leaves one possible outcome for the remaining values, the final unique solution below:
As an unrelated statistical challenge, what was the probability of Felix achieving this scoreboard? Assume, of course, fair dice.
Game 1 = 6! × (5C5·50·6−5) × (5C2·53·6−5) × (5C2·53·6−5) × (5C2·53·6−5) × (5C2·53·6−5) × (5C5·50·6−5)
Game 2 = 6! × (5C1·54·6−5) × (5C4·51·6−5) × (5C3·52·6−5) × (5C4·51·6−5) × (5C1·54·6−5) × (5C4·51·6−5)
Game 3 = 6! × (5C4·51·6−5) × (5C1·54·6−5) × (5C5·50·6−5) × (5C1·54·6−5) × (5C4·51·6−5) × (5C3·52·6−5)
Game 4 = 6! × (5C3·52·6−5) × (5C5·50·6−5) × (5C1·54·6−5) × (5C5·50·6−5) × (5C3·52·6−5) × (5C2·53·6−5)
Game 5 = 6! × (5C2·53·6−5) × (5C3·52·6−5) × (5C4·51·6−5) × (5C3·52·6−5) × (5C5·50·6−5) × (5C1·54·6−5)
Total = (6!)5 × (5C1·5C2·5C3·5C4·5C5)6 × 560 × 6−150 = 565 × 3−140 × 2−118 = 1.30177...×10−57
The basic numerical solution (assuming random allocation) is shown above, but how does a strategy of choice affect this value?